[Physics] From uncertainty to commutation relations

commutatorheisenberg-uncertainty-principlemeasurement-problemobservablesquantum mechanics

Consider the famous problem of measuring both the position and momentum of an electron. We start with two photographs at different times, then worry about the momentum of the photon, the wavelength of the light, the aperture of the camera etc. We end up with the classical uncertainty relations; well actually a bit more: we have an idea of the signs of the errors (e.g. we know on which side the electron was hit by the photon).

Now suppose that we have an idea that the measurements are made by application of SOME operators (not necessarily the standard ones) on a Hilbert space. Can we deduce anything like the canonical commutation relations between position and momentum operators? In other words, is the implication between commutation and order of measurement only one way (the commutation relations are consistent with the measurements), or can we give a limited implication the other way?

The interest is with novel quantum systems – if there is an idea how measurement theory works, then is this any guide to the algebra of observables? (As well as being of general historical or philosophical interest in whether quantum theory might have taken another turn using different operator representations.)

Best Answer

The uncertainty principle between two observables is related to their commutator in a general and profund way. The generalized uncertainty principle can be proved quite generally using simple matrix algebra and the Cauchy-Schwartz Inequality:

I) Supose we have two hermitian operators (aka observables) $\hat{A}$ and $\hat{B}$. The possible results of a measurment are their eivenvalues and the dispersion in the measurment is:

\begin{equation} ({\Delta\hat{A}})^2 = \langle\hat{A}^2\rangle-\langle\hat{A}\rangle^2 \end{equation}

We can calways chose a new reference system to set $<\hat{A}>=0$ so we get:

\begin{equation} ({\Delta\hat{A}})^2 = \langle\hat{A}^2\rangle = \int\psi^*\hat{A}^2\psi dx =\langle\psi|\hat{A}^2|\psi\rangle \end{equation}

And obviusly the same holds for $\hat{B}$.

II) Using the Cauchy-Schwartz Inequality:

\begin{equation} \langle\psi|\hat{A}\hat{A}|\psi\rangle\langle\psi|\hat{B}\hat{B}|\psi\rangle \geqslant |\langle\psi|\hat{A}\hat{B}|\psi\rangle|^2 \end{equation}

One can inmediately obtain:

\begin{equation} ({\Delta\hat{A}})^2({\Delta\hat{B}})^2 \geqslant |\langle\psi|\hat{A}\hat{B}|\psi\rangle|^2 \end{equation}

III) Now,we can reduce the term in the right

\begin{equation} |\langle\psi|\hat{A}\hat{B}|\psi\rangle| \geqslant |Im\Big[\langle\psi|\hat{A}\hat{B}|\psi\rangle \Big] = |\frac{1}{2i}\Big[\langle\psi|\hat{A}\hat{B}|\psi\rangle - \langle\psi|\hat{A}\hat{B}|\psi\rangle^* \Big]| \end{equation}

Where I have used that the modulus of a complex number is bigger than its Imaginary part and then I used the fact that if $f= Re(f)+i Im(f)$ then $Im(f)=\frac{1}{2i}(f-f^*)$.

IV) Because $\hat{A}$ and $\hat{B}$ are observables then $\langle\psi|\hat{A}\hat{B}|\psi\rangle^*=\langle\psi|(\hat{A}\hat{B})^{\dagger}|\psi\rangle=\langle\psi|(\hat{B}\hat{A})|\psi\rangle$

V) Finally, using this result we can rewrite the inequality as:

\begin{equation} ({\Delta\hat{A}})^2({\Delta\hat{B}})^2 \geqslant |\frac{1}{2i}\Big[\langle\psi|\hat{A}\hat{B}|\psi\rangle - \langle\psi|\hat{B}\hat{A}|\psi\rangle \Big]| =|\frac{1}{2i}\Big[\langle\hat{A}\hat{B}\rangle - \langle\hat{B}\hat{A}\rangle \Big] |=|\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle| \end{equation}

So the dispersion in any two hermitian operators is related to their commutator

\begin{equation} ({\Delta\hat{A}})^2({\Delta\hat{B}})^2 \geqslant|\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle| \end{equation}

I suppose that you could measure the dispersion of two observables with increasing accuracy to find some upper limmit on their commutator.

Note that this work for any two observables you like to use, not just canonical ones like X and P.

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